Switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$

Question

How could I switch from $a\cdot \sin(t) + b \cdot \cos(t)$ to $c \cdot \cos(t+f)$? Thank you for your time.

Answer 1

We see that by the addition formula,

$$c \cdot \cos(t + f) = c \cdot \left( \cos(t)\cos(f) - \sin(t)\sin(f)\right).$$

If you set $a = -\sin(f), c = 1,$ and $b = \cos(f),$ You can make a reasonable comparison between the expressions $a \sin(t) + b \cos(t)$ and $c \cdot \cos(t + f).$

Answer 2

$$ a\sin t+b\cos t=\sqrt{a^2+b^2}\left[\frac{b}{\sqrt{a^2+b^2}}\cos t +\frac{a}{\sqrt{a^2+b^2}}\sin t\right]$$

$$ =\sqrt{a^2+b^2}\left[\cos \alpha \cos t +\sin\alpha \sin t\right]$$

$$ =\sqrt{a^2+b^2} \cos(t-\alpha). $$

$ (c,f) $ have different symbols.

Useful in combining two waves of same frequency but different amplitude.

Answer 3

Set $t=0$, giving

$$a\sin(0)+b\cos(0)=b=c\cos(f).$$

Then set $t=\frac\pi2$, and

$$a\sin\left(\frac\pi2\right)+b\cos\left(\frac\pi2\right)=a=c\cos\left(\frac\pi2+f\right)=-c\sin(f).$$

This gives you the transfrom from $c,f$ to $a,b$.

You can invert it by noting that

$$a^2+b^2=c^2\cos^2(f)+c^2\sin^2(f)=c^2,$$ and

$$\frac ab=-\frac{c\sin(f)}{c\cos(f)}=-\tan(f).$$

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Alternatively, the left expression cancels when

$$a\sin(t)+b\cos(t)=0,$$ i.e.

$$\tan(t)=-\frac ba.$$

This must coincide with

$$\cos(t+f)=0,$$ or

$$f=\frac\pi2-t,$$

$$\tan(f)=\cot(t)=-\frac ab.$$

Then, with $t=-f$,

$$-a\sin(f)+b\cos(f)=c,$$

$$c=-a\sin\left(\arctan\left(-\frac ab\right)\right)+b\cos\left(\arctan\left(-\frac ab\right)\right).$$